How Do You Calculate Work Done by a Force Field Along Different Paths?

In summary, the conversation discusses a problem involving a force that decreases with F=(1/r^2)r, where F is a vector and r is a unit vector. The speaker needs to find the work done by a direct radial path from r=sqrt(2) to r=2*sqrt(2), as well as a path from (1,1) to (2,1) to (2,2). The solution involves calculating integrals for each component of the vector and using parameter equations. The speaker is able to complete the problem with the help of the expert.
  • #1
ossito_the-diracian
4
0
My problem has force decreasing with F=(1/r^2)r, where F is a vecotr and r is unit vector. i need to find a).work done in moving from a point at r=sqrt(2) to a point at r=2*sqrt(2) by a direct radial path and (b) by a path from (1,1)-->(2,1)-->(2,2). Compare my answers.

a)I did direct radial path using Work=Integral[1/r^2] from sqrt(2) to 2*sqrt(2). I got sqrt(2)/2.

b) This where problem is: from (1,1)-->(2,1) x:1-->2, y=1, dl=dx x, so i get F (dot) dl = (x^2+y^2)^2, however, I am not sure I am setting up the x and y components correctly. from (2,1)-->(2,2) y:1-->2, x=2 dl=dy y and again same issues of x and y components.

i did read that r vector=sin(theta)cos(phi)x+sin(theta)sin(phi)y, but was not sure how to incorparate this into line integral part of problem.


any help would be appreciated, thanks
 
Physics news on Phys.org
  • #2
First of all this is a vector-integral. This means that you are going to have to calculate this integral for each component. In your case we have two components x and y so r²=x² + y².

Now (1/r²)rdr = (1/r²)dx + (1/r²)dy. Because the unitvector multiplied by the vector dr gives you dx and dy (via the dot-product)

Now when you move from (1,1) to (2,1) you will need a parameter-equation for the integration. y = 1 and x=l with l the parameter and l : 1-->2. So in the integral replace x by l and dx=dl and y = 1 and dy =0

The x-component of the integration is integral(dl/(l²+1²)) and l starts in 1 and stops in 2. this integral is easy because you know that generally
integral(1/(1 + x²)dx) yields arctan(x).

Since along this path dy = 0 the second term of the integral vanishes...
So the only thing you need to do now is calculate the integral correctly. But the solution of this integral will be an arcus-tangens...


The other path is calculated in the exact same way.


regards
marlon
 
  • #3
thanks a bunch, I was sort of close, but never looked at the vector part of it. My first integral where dy=0 I was using (x^2+1)dx from 1-->2. I should be able to complete it now, again thank you.
 

1. What is an E&M line integral setup?

An E&M (electromagnetism) line integral setup is a mathematical tool used in physics to calculate the work done by an electric or magnetic field along a specific path. It involves taking a line integral of a vector field over a given curve.

2. How is an E&M line integral setup different from a regular line integral?

An E&M line integral setup is specific to electric and magnetic fields, while a regular line integral can be used for any vector field. The difference lies in the integrand, which is a vector function in E&M line integrals and a scalar function in regular line integrals.

3. What are the applications of E&M line integral setups?

E&M line integral setups are commonly used in physics and engineering to calculate the work done by electric and magnetic fields. They are also useful in calculating the flux of a vector field and determining the potential energy of a system.

4. How do you set up an E&M line integral?

To set up an E&M line integral, you first need to determine the vector field and the curve over which the line integral will be calculated. Then, you need to parameterize the curve and find the dot product between the vector field and the tangent vector of the curve. Finally, integrate the resulting expression over the curve.

5. Are there any limitations to using E&M line integral setups?

One limitation of E&M line integral setups is that they can only be used for conservative vector fields, meaning that the work done along a closed path is zero. They also cannot be used for non-conservative fields, such as those with sources or sinks.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
866
  • Introductory Physics Homework Help
2
Replies
64
Views
2K
  • Introductory Physics Homework Help
Replies
25
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
915
  • Introductory Physics Homework Help
Replies
1
Views
191
Replies
8
Views
228
  • Introductory Physics Homework Help
Replies
1
Views
882
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
1K
Back
Top